[/quote]
,
and 
be two fixed points in the plane. For each point 
of the plane, outside of the line 
,
be the barycenter of the triangle 
.
.
be a tree on 
vertices with exactly 
leaves. Suppose that there exists a subset of at least 
vertices of 
be an acute scalene triangle. Denote by 
the circle with diameter 
be the contact points of the tangents from 
lie on opposite sides of 
and 
lie on opposite sides of 
.
be the circle with diameter 
,
,
the circle with diameter 
.
are concurrent.
intersects the line 
at 
and 
is the midpoint of 
.
and 
intersect on 
![\[x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0\]](http://latex.artofproblemsolving.com/b/9/d/b9d355433e896e8b97f28c2e4235140fd2348e77.png)
correct to four decimal places.
be an integer, 
a real number, and 
be positive real numbers such that 
.![\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]](http://latex.artofproblemsolving.com/0/8/d/08de0c74a4e36b50a64d17875d3fd93eeb5b52de.png)
be positive real numbers. Prove that there exists a cyclic permutation 
of 
such that for all positive real numbers 
the following holds:![\[
\frac{a}{x\,a + y\,b + z\,c}
\;+\;
\frac{b}{x\,b + y\,c + z\,a}
\;+\;
\frac{c}{x\,c + y\,a + z\,b}
\;\ge\;
\frac{3}{x + y + z}.
\]](http://latex.artofproblemsolving.com/1/5/2/15261b3f712c2cc8ffbeb92a9711e06ad0992b5d.png)
be positive integers such that 
,
,
are perfect squares. Prove that 
is also a perfect square.
be distinct primes and let 
be nonnegative integers. Define![\[
m \;=\;
\frac12
\Bigl(\prod_{i=2}^k p_i^{a_i}\Bigr)
\Bigl(\prod_{i=1}^k(p_i+1)\;+\;\sum_{i=1}^k(p_i-1)\Bigr),
\]](http://latex.artofproblemsolving.com/7/9/a/79af85375957986298d37c00d288b6d402f40106.png)
![\[
n \;=\;
\frac12
\Bigl(\prod_{i=2}^k p_i^{a_i}\Bigr)
\Bigl(\prod_{i=1}^k(p_i+1)\;-\;\sum_{i=1}^k(p_i-1)\Bigr).
\]](http://latex.artofproblemsolving.com/e/2/3/e23799ff03400d5eade3731bb83456fcd702a5e1.png)
Prove that![\[
p^2-1 \;\bigm|\; p\,m \;-\; n.
\]](http://latex.artofproblemsolving.com/3/7/e/37e4535b59ef183908730946213429e9ebc05dd7.png)
,
.
and 
such that the only 
equals its 
-
.
digit numbers 
(with 
)
,
,
.
be the midpoint of 
.
.
,
in degrees?
with 
and 
for 
.
.
.
.
(x) be the 
.
such that 
divides 
?
be the set of vectors 
where 
are all real numbers. Let 
denote 
.
be the set in 
given by 
If a point 
is uniformly at random from ![$E[||z||^2]$](http://latex.artofproblemsolving.com/e/1/a/e1aa966627c05d1f317981d82094e07d1b1ce0e1.png)
?
be the unique integer between 
and 
,
.
and 
,
.
,
,
,
.
and 
,
and the puck bounces perfectly around the box (angle in equals angle out, no friction) until it hits one of the four vertices of the box. What is the expected number of times it will hit an edge or vertex of the box, including both when it starts at 
.
,
,
.
graph paper notebooks. How many ways can she buy school supplies?
be the center of the circumcircle of right triangle 
.
be the midpoint of minor arc 
be a point on line 
.
be the intersection of line 
and the Circle 
be the intersection of line 
and 
.
and 
,
.
to be the maximum possible least-common-multiple of any sequence of positive integers which sum to 
be positive reals such that 
.
![\[
M = \frac{1 - x^2}{x^2} + 2y^2 + 3x + \frac{24}{y} + 2025.
\]](http://latex.artofproblemsolving.com/6/b/d/6bd846ab0f95b1cf4766c7bf0356a439f3bfc93a.png)
Y by Adventure10
One base of a trapezoid is 100 units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio 
Let 
be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed 
Let 
be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed 
,
.
.
,
.
as the height of 
as base, and the length of the parallel segment as 
.
.
and 
.
.
.
,
.
,
.
,
.
,
.
and 
.
,
on 
,
.![$\frac{[ABFE]}{[CDFE]}=\frac{AB+25}{AB+75}=\frac23$](http://latex.artofproblemsolving.com/0/1/b/01b532d4b4a9f998197e484bbcf063cdb89de4eb.png)
,
and 
.
,
.
.
,
.
,
.
and 
be the segment that divides the trapezoid into equal areas. 
,
.![$[EAB]: [ABQP]: [QPCD]=9: 20: 20$](http://latex.artofproblemsolving.com/7/a/8/7a8b057a72537b62a302223f72aa4e1a54901043.png)
,
.
,
.
.
?
?
be the lengths of the bases. We have that the midsegment has length 
,![\[\frac 23 = \frac{a+a+50}{a+100+a+50}\implies a = 75.\]](http://latex.artofproblemsolving.com/4/d/2/4d26a10dc1bc3478e24e4ec82ac0888dc792ae1c.png)
Now, it is well-known that the area of the segment dividing the trapezoid into equal segments has length 
(proof: intersect the legs at a point and use similar triangles) so the answer is![\[\left\lfloor\frac{75^2+175^2}{200}\right\rfloor = \left\lfloor\frac{15^2+35^2}{8}\right\rfloor = \left\lfloor\frac{225+1225}{8}\right\rfloor = \left\lfloor\frac{1450}{8}\right\rfloor=181.\]](http://latex.artofproblemsolving.com/4/c/d/4cd2fd04322c33d30802569b2a1c291a2e574034.png)
,
on 
.
.
.
